# Difference of Lebesgue-measurable sets [duplicate]

Let $$A, B\subset\mathbb{R}$$ two Lebesgue-measurable sets of positive Lebesgue-measure. Prove that $$A-B$$ contains an interval.

I thought that being the measure positive then the Hausdorff dimension of the two sets is 1. But I don't know how to go further. The hint says: "use the convolution of the two characteristics function of the two sets" but I can't really figure out how to use it.

It is enough to consider the case when $$A$$ and $$B$$ are bounded. Let $$f=\chi_A * \chi_B$$. It is a general fact that convolution of two functions in $$L^{1} \cap L^{\infty}$$ is continuous. [This is proved by approximating these functions by continuous with compact support in $$L^{1}$$ norm]. Now $$\int f(x)dx=\int \chi_A(x)dx \int \chi_B(x)dx=m(A)m(B) >0$$. Hence $$f(x_0) >0$$ for some $$x_0$$ and there exists $$\epsilon >0$$ such that $$f(x) >0$$ for all $$x \in (x_0-\epsilon ,x_0+ \epsilon)$$. So for any $$x$$ in this interval there exists $$y$$ such that $$\chi_A(x-y)\chi_B(y) >0$$ which means $$y \in B \cap (x-A)$$. It follows that $$x =(x-y)+y \in A+B$$.
We have proved that $$A+B$$ contains an interval. For $$A-B$$ just change $$B$$ to $$-B$$.